Consider the real projective plane $\mathbb{RP}^2$ as a metric space, with the distance between two points $[x_0:x_1:x_2]$ and $[y_0:y_1:y_2]$ given by the angle between the lines through the origin in $\mathbb{R}^3$ that the vectors $(x_0,x_1,x_2)$ and $(y_0,y_1,y_2)$ define.
A function $f: \mathbb{RP}^2 \to \mathbb{RP}^2$ is called a projective transformation is there exists some bijective linear map $T: \mathbb{R}^3 \to \mathbb{R}^3$ which induces it, i.e., $f([x_0:x_1:x_2])=[T(x_0,x_1,x_2)]$.
How can one prove that every isometry of $\mathbb{RP}^2$ is a projective transformation? I know it's true, but I wasn't able to prove it or find a proof of this anywhere online.
One approach I tried was proving that an isometry of $\mathbb{RP}^2$ is uniquely determined given the images of a projective frame.
For that I tried to prove that a point of $\mathbb{RP}^2$ can be uniquely determined given its distances to the points of a projective frame, but it turns out that is false: if we consider the canonical projective frame ($[1:0:0],[0:1:0],[0:0:1],[1:1:1]$), the points $[1:-1:1]$ and $[-1:1:1]$ have the same distances to the points of the frame. Is this a good approach or is it a dead end?